An electrolytic cell contains a solution of `Ag_(2)SO_(4)` and have platinum electrodes. A current is passed until 1.6gm of `O_(2)` has been liberated at anode. The amount of silver deposited at cathode would be

Options

A. 107.88 gm
B. 1.6 gm
C. 0.8 gm
D. 21.60 gm

Answer

D

Explanation

At cathode: `Ag^(+)+e^(-)toAg`
At anode: `2OH^(-)tH_(2)O+(1)/(2)O_(2)+2e^(-)`
`E_(Ag)=(108)/(1)=108,E_(O_(2))=((1)/(2)xx32)/(2)=8`
`(W_(Ag))/(E_(Ag))=(W_(O_(2)))/(E_(O_(2))),W_(Ag)=(1.6xx108)/(8)=21.6gm`